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Chapter 12: Problem 16

\(\boldsymbol{F}(\boldsymbol{x}, \boldsymbol{y}, z)=\boldsymbol{0}\) Find an equation of the plane tangent to the following surfaces at the given points (two planes and two equations). $$2 x+y^{2}-z^{2}=0 ;(0,1,1) \text { and }(4,1,-3)$$

Short Answer

Expert verified
The equation of the tangent plane at point (0,1,1) is: $$x + y - z = 0$$. The equation of the tangent plane at point (4,1,-3) is: $$x + y + 3z = 13$$.

Step by step solution

01

Find the gradient of the surface

We start by finding the gradient of the given surface, which is represented as a vector. The surface equation is $$2x+y^2-z^2=0$$. We need to find partial derivatives of the equation with respect to x, y, and z: $$ \frac{\partial}{\partial x}(2x+y^2-z^2) = 2 \\ \frac{\partial}{\partial y}(2x+y^2-z^2) = 2y \\ \frac{\partial}{\partial z}(2x+y^2-z^2) = -2z $$ The gradient vector at point \((x, y, z)\) is then given by: $$\nabla f(x, y, z) = \langle 2, 2y, -2z \rangle$$
02

Find the gradient vectors at the given points

Now, we need to find the gradient vectors at the given points (0,1,1) and (4,1,-3). Using the gradient vector formula from Step 1: $$\nabla f(0, 1, 1) = \langle 2, 2(1), -2(1) \rangle = \langle 2, 2, -2 \rangle$$ $$\nabla f(4, 1, -3) = \langle 2, 2(1), -2(-3) \rangle = \langle 2, 2, 6 \rangle$$
03

Use the gradient vectors and given points to find the plane equation

The tangent planes at the given points have normal vectors equal to the respective gradient vectors. Using the normal vector and the point, we can determine the equation of the tangent planes at each point. For the tangent plane at point (0,1,1), using the gradient vector \(\langle 2, 2, -2 \rangle\) and the point-plane equation form: $$2(x - 0) + 2(y - 1) - 2(z - 1) = 0$$ Simplifying, we get the tangent plane equation at point (0,1,1): $$x+ y-z = 0$$ For the tangent plane at point (4,1,-3), using the gradient vector \(\langle 2, 2, 6 \rangle\) and the point-plane equation form: $$2(x - 4) + 2(y - 1) + 6(z + 3) = 0$$ Simplifying, we get the tangent plane equation at point (4,1,-3): $$x+y+3z = 13$$
04

Final Answer

The equation of the tangent plane at point (0,1,1) is: $$x + y - z = 0$$. The equation of the tangent plane at point (4,1,-3) is: $$x + y + 3z = 13$$.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Partial Derivatives
In calculus, partial derivatives represent how a function changes as one specific variable changes, while all other variables are held constant. Consider the function given by the equation of the surface: \(2x+y^2-z^2=0\). To find the partial derivatives, you differentiate with respect to one variable at a time:
  • With respect to \(x\): The derivative is \(2\), since \(2x\) is linear in \(x\) and the other terms are constants.
  • With respect to \(y\): The derivative is \(2y\), as only the term \(y^2\) involves \(y\).
  • With respect to \(z\): The derivative is \(-2z\), since only \(z^2\) involves \(z\), and the negative sign is because of the \(-z^2\) term.
The partial derivatives are crucial because they form the components of the gradient vector, indicating the slope or rate of change in the direction of each axis.
Understanding these derivatives helps in constructing more complex concepts like the gradient vector, which we will explore next.
Gradient Vector
The gradient vector, often denoted as \(abla f\), combines all partial derivatives into a single vector. It's an essential tool in multivariable calculus as it points in the direction of the steepest ascent of the function. For the given surface \(2x+y^2-z^2=0\), the gradient vector is:
  • \(abla f(x, y, z) = \langle 2, 2y, -2z \rangle\)
This vector represents the rate of change of the function in \(x\), \(y\), and \(z\) directions.
At a specific point, like \((0, 1, 1)\), we evaluate the gradient vector by substituting the coordinates into its components:
  • \(abla f(0, 1, 1) = \langle 2, 2, -2 \rangle\)
  • \(abla f(4, 1, -3) = \langle 2, 2, 6 \rangle\)
The evaluated gradient at these points acts as a normal vector for the tangent plane at the point on the surface. The gradient vector's direction and magnitude can give insights into how steep or gradual the surface changes over the defined point.
Normal Vector
In the context of tangent planes, the normal vector is a vector perpendicular to the plane. In this exercise, the gradient vector also plays the role of the normal vector, which helps determine the orientation of the tangent plane.
Using the given surface equation \(2x+y^2-z^2=0\), we compute the gradient vectors at specific points:
  • At point \((0,1,1)\), the normal vector is \(\langle 2, 2, -2 \rangle\).
  • At point \((4,1,-3)\), the normal vector is \(\langle 2, 2, 6 \rangle\).
These vectors are fundamental in establishing the equation of the tangent plane.
The idea here is that every plane at a point on the surface shares a perpendicular direction with its normal vector, ensuring that the plane is tangential at just that particular point on the surface. Understanding normal vectors is key for comprehending concepts in three-dimensional space geometry and calculus.
Surface Equation
The surface equation forms the foundation upon which tangent planes are determined. In this exercise, the surface is defined by \(2x+y^2-z^2=0\). This equation describes a three-dimensional shape in space, specifying the relationship between the variables \(x\), \(y\), and \(z\).
From this surface equation, important details such as the gradient and normal vectors are derived. This makes it possible to compute the equation for tangent planes at specific points on the surface:
  • For point \((0, 1, 1)\), the tangent plane equation simplifies to \(x+y-z=0\).
  • For point \((4, 1, -3)\), the tangent plane equation becomes \(x+y+3z=13\).
These equations describe planes that just touch the surface at the respective points. They are crucial in fields requiring precise calculations for how surfaces interact with surrounding space, like in physics and engineering.
Understanding how to derive tangent planes from a surface equation enhances problem-solving skills in multivariable calculus and its applications.

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Most popular questions from this chapter

Given three distinct noncollinear points \(A, B,\) and \(C\) in the plane, find the point \(P\) in the plane such that the sum of the distances \(|A P|+|B P|+|C P|\) is a minimum. Here is how to proceed with three points, assuming that the triangle formed by the three points has no angle greater than \(2 \pi / 3\left(120^{\circ}\right)\). a. Assume the coordinates of the three given points are \(A\left(x_{1}, y_{1}\right)\) \(\underline{B}\left(x_{2}, y_{2}\right),\) and \(C\left(x_{3}, y_{3}\right) .\) Let \(d_{1}(x, y)\) be the distance between \(A\left(x_{1}, y_{1}\right)\) and a variable point \(P(x, y) .\) Compute the gradient of \(d_{1}\) and show that it is a unit vector pointing along the line between the two points. b. Define \(d_{2}\) and \(d_{3}\) in a similar way and show that \(\nabla d_{2}\) and \(\nabla d_{3}\) are also unit vectors in the direction of the line between the two points. c. The goal is to minimize \(f(x, y)=d_{1}+d_{2}+d_{3} .\) Show that the condition \(f_{x}=f_{y}=0\) implies that \(\nabla d_{1}+\nabla d_{2}+\nabla d_{3}=0\) d. Explain why part (c) implies that the optimal point \(P\) has the property that the three line segments \(A P, B P,\) and \(C P\) all intersect symmetrically in angles of \(2 \pi / 3\) e. What is the optimal solution if one of the angles in the triangle is greater than \(2 \pi / 3\) (just draw a picture)? f. Estimate the Steiner point for the three points (0,0),(0,1) and (2,0).

Identify and briefly describe the surfaces defined by the following equations. $$x^{2} / 4+y^{2}-2 x-10 y-z^{2}+41=0$$

Extending Exercise \(62,\) when three electrical resistors with resistance \(R_{1}>0, R_{2}>0,\) and \(R_{3}>0\) are wired in parallel in a circuit (see figure), the combined resistance \(R,\) measured in ohms \((\Omega),\) is given by \(\frac{1}{R}=\frac{1}{R_{1}}+\frac{1}{R_{2}}+\frac{1}{R_{3}}\) Estimate the change in \(R\) if \(R_{1}\) increases from \(2 \Omega\) to \(2.05 \Omega, R_{2}\) decreases from \(3 \Omega\) to \(2.95 \Omega,\) and \(R_{3}\) increases from \(1.5 \Omega\) to \(1.55 \Omega.\)

Traveling waves (for example, water waves or electromagnetic waves) exhibit periodic motion in both time and position. In one dimension, some types of wave motion are governed by the one-dimensional wave equation $$\frac{\partial^{2} u}{\partial t^{2}}=c^{2} \frac{\partial^{2} u}{\partial x^{2}},$$ where \(u(x, t)\) is the height or displacement of the wave surface at position \(x\) and time \(t,\) and \(c\) is the constant speed of the wave. Show that the following functions are solutions of the wave equation. $$u(x, t)=5 \cos (2(x+c t))+3 \sin (x-c t)$$

Given a differentiable function \(w=f(x, y, z),\) the goal is to find its maximum and minimum values subject to the constraints \(g(x, y, z)=0\) and \(h(x, y, z)=0\) where \(g\) and \(h\) are also differentiable. a. Imagine a level surface of the function \(f\) and the constraint surfaces \(g(x, y, z)=0\) and \(h(x, y, z)=0 .\) Note that \(g\) and \(h\) intersect (in general) in a curve \(C\) on which maximum and minimum values of \(f\) must be found. Explain why \(\nabla g\) and \(\nabla h\) are orthogonal to their respective surfaces. b. Explain why \(\nabla f\) lies in the plane formed by \(\nabla g\) and \(\nabla h\) at a point of \(C\) where \(f\) has a maximum or minimum value. c. Explain why part (b) implies that \(\nabla f=\lambda \nabla g+\mu \nabla h\) at a point of \(C\) where \(f\) has a maximum or minimum value, where \(\lambda\) and \(\mu\) (the Lagrange multipliers) are real numbers. d. Conclude from part (c) that the equations that must be solved for maximum or minimum values of \(f\) subject to two constraints are \(\nabla f=\lambda \nabla g+\mu \nabla h, g(x, y, z)=0\) and \(h(x, y, z)=0\).
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